CF825D：Suitable Replacement（字符串）

最新推荐文章于 2025-12-06 13:16:38 发布

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本文介绍了一个字符串处理问题，目标是通过替换特殊字符并重新排列字符串s中的字母，使其包含尽可能多的非交叠子串t。文章提供了一种解决思路，采用二分查找法确定最优的子串匹配数量，并附带了完整的C++代码实现。

You are given two strings s and t consisting of small Latin letters, string s can also contain '?' characters.

Suitability of string s is calculated by following metric:

Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings s, you choose the one with the largest number of non-intersecting occurrences of string t. Suitability is this number of occurrences.

You should replace all '?' characters with small Latin letters in such a way that the suitability of string s is maximal.

Input

The first line contains string s (1 ≤ |s| ≤ 106).

The second line contains string t (1 ≤ |t| ≤ 106).

Output

Print string s with '?' replaced with small Latin letters in such a way that suitability of that string is maximal.

If there are multiple strings with maximal suitability then print any of them.

Examples

input
?aa?
ab

output
baab

input
??b?
za

output
azbz

input
abcd
abacaba

output
abcd

Note

In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2.

In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them.

In the third example there are no '?' characters and the suitability of the string is 0.

题意：给出字符串s和t，s可以任意调转字符顺序，要求将问号替换成字符，使得s中包含最多的不相交的t串。

思路：二分答案，设p为s中出现的t数量，二分p即可。

# include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
char s[maxn], t[maxn], st[maxn];
int w=0, cnt=0, T[28]={0}, S[28]={0};
int slen, tlen;
bool check(int x)
{
    int sum = 0;
    for(int i=0; i<26; ++i) sum = sum + max(0, T[i]*x-S[i]);
    return sum <= w;
}
int main()
{
    scanf("%s%s",s+1,t+1);
    slen = strlen(s+1), tlen = strlen(t+1);
    int p = slen/tlen;
    for(int i=1; t[i]; ++i) ++T[t[i]-'a'];
    for(int i=1; s[i]; ++i)
    {
        if(s[i]=='?') ++w;
        else ++S[s[i]-'a'];
    }
    int l=0, r=p;
    while(l<=r)
    {
        int m = l+r>>1;
        if(check(m)) l=m+1;
        else r=m-1;
    }
    for(int i=0; i<26; ++i)
    {
        int tmp = max(0, T[i]*r-S[i]);
        while(tmp--) st[cnt++] = i+'a';
    }
    for(int i=1; i<=slen; ++i)
    {
        if(s[i]=='?')
        {
            if(cnt) printf("%c",st[--cnt]);
            else printf("a");
        }
        else
            printf("%c",s[i]);
    }
    return 0;
}